R help - Mar 2009 - loop problem

Dear R members
I have a problem regarding storing the lists.
Let 
L=number of distinct values of any predictor (say L=5)
P=number of predictors (say P=20)

g1 <- c()
for(i in 1:P){
if(L > 1){
  for(j in 1:(L-1)){
g <- ....
g1[j] <- g
                       }
            }
g2[]=sort.list(g1)
                }

Now the question is: What should we use inside brackets of g2[....], whether
"i" or some thing else? If L is not greater than 1 then there will be
a "NULL" for g2. We don't want to store it in g2, so how can we
handle this problem. Looking forward for some help. Thanks and

 
best regards

Muhammad Azam 



      
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Muhammad Azam wrote:
> Dear R members
> I have a problem regarding storing the lists.
> Let 
> L=number of distinct values of any predictor (say L=5)
> P=number of predictors (say P=20)
>
> g1 <- c()
> for(i in 1:P){
> if(L > 1){
>   for(j in 1:(L-1)){
> g <- ....
> g1[j] <- g
>                        }
>             }
> g2[]=sort.list(g1)
>                 }
>
> Now the question is: What should we use inside brackets of g2[....],
whether "i" or some thing else? If L is not greater than 1 then there
will be a "NULL" for g2. We don't want to store it in g2, so how
can we handle this problem. Looking forward for some help. Thanks and
>   
Hi Muhammad,
The first thing I would ask is "Why not store NULL in g2[[i]] if that 
element is NULL?". If you do this:

for(i in 1:P) {
 if(L > 1) {
  g1<-rep(NA,L)
  for(j in 1:(L-1)) g1[j]<-...
  g2[[i]]<-sort.list(g1)
 }
}

you will have fewer elements in g2 than you have sets of predictors and 
will then have to match up the sets of predictors with the elements in 
g2. I think you could just test for NULL

is.null(g2[[i]])

later on if you don't want to process these in a subsequent step.

Jim